Time and Work Problems: Method with Worked Examples

The steady-rate problems below look different from each other but reduce to one method: turn each time into a rate, combine the rates (adding them, or subtracting one that works against the rest, like an outlet), and invert. Five worked examples follow, covering two workers, three workers, an inlet-outlet pair and an efficiency ratio. Every answer is exact arithmetic under two stated assumptions: each worker or pipe keeps a constant rate, and the rates combine without interference.

The method

If someone finishes a job in d days, they complete 1/d of the job per day. When people work together, add their per-day rates; when something works against the job, subtract its rate. If the net rate is r of the job per day, the time to finish the whole job is T = 1/r days, provided r is positive. For two people who take a and b days alone, the combined time is ab/(a + b) days.

d

days one worker needs to finish the whole job alone

1/d

that worker's rate: fraction of the job done per day

T

days needed when working together

Worked examples

Example 1: two workers

A can finish a job in 6 days and B in 12 days. How long do they take together?

1. Rates: A does 1/6 of the job per day; B does 1/12.
2. Combined rate: 1/6 + 1/12 = 3/12 = 1/4 of the job per day.
3. Time together: T = 1/(1/4) = 4 days.

Together they take 4 days.

Example 2: a less tidy pair

A takes 10 days and B takes 15 days. How long together?

1. Combined rate: 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 of the job per day.
2. Time together: T = 6 days.

Together they take 6 days.

Example 3: three workers

A, B and C alone take 4, 6 and 12 days. How long do they take together?

1. Combined rate: 1/4 + 1/6 + 1/12 = 3/12 + 2/12 + 1/12 = 6/12 = 1/2 of the job per day.
2. Time together: T = 2 days.

Together they take 2 days.

Example 4: inlet and outlet

An inlet pipe fills an initially empty tank in 6 hours at a constant flow. An outlet pipe empties a full tank in 12 hours at a constant flow. With both open, how long does the empty tank take to fill?

1. The outlet works against the inlet, so subtract its rate: 1/6 - 1/12 = 2/12 - 1/12 = 1/12 of the tank per hour.
2. The net rate 1/12 is positive, so the tank does fill, in T = 12 hours.

The tank fills in 12 hours.

Example 5: efficiency ratio

A is twice as fast as B, and B alone takes 18 days. How long do they take together?

1. Twice as fast means half the time: A alone takes 9 days.
2. Combined rate: 1/9 + 1/18 = 2/18 + 1/18 = 3/18 = 1/6 of the job per day.
3. Time together: T = 6 days.

Together they take 6 days.

A staged-work variant

A takes 8 days and B takes 12 days. A works alone for 2 days, then B joins. A completes 2/8 = 1/4 of the job first, leaving 3/4. The combined rate is 1/8 + 1/12 = 5/24 of the job per day, so the rest takes (3/4)/(5/24) = 18/5 = 3.6 days. Total time: 2 + 18/5 = 28/5 = 5.6 days. Keep fractions until the final step to avoid rounding errors.

Frequently asked questions

How do I solve a time and work problem?

Convert each completion time into a rate, such as 1/d of the job per day. Add rates that help and subtract rates that oppose the job. If the net rate r is positive, the time is 1/r. This assumes constant, additive rates; a zero or negative net rate cannot finish the job.

What is the formula for two people working together?

If A alone takes a days and B alone takes b days, their rates add: 1/a + 1/b of the job per day, so together they take ab/(a + b) days. The formula holds only under the stated assumption of constant rates that add, with no interference between the workers.

How is an inlet and outlet (pipes) problem different?

An outlet works against the filling, so its rate is subtracted, not added. For an initially empty tank with constant flows, an inlet at 1/6 per hour and an outlet at 1/12 per hour give a net 1/12 per hour, which is positive, so the tank fills in 12 hours.

What does 'A is twice as fast as B' mean in days?

Twice as fast means half the time for the same job: if B takes 18 days, A takes 9 days. Rates, not times, are what add: 1/9 + 1/18 = 1/6, so working together they take 6 days under the same steady-rate assumption.

Do these results predict real work?

No. They are exact arithmetic under the stated assumptions of constant, additive rates. Real work varies with fatigue, skill and coordination, so read every answer as the model's answer, not a prediction about real people or projects.

Sources and scope

Method checked against Purplemath, "Work" Word Problems, and OpenStax, Elementary Algebra 2e, section 8.8, both retrieved 15 November 2026. This article teaches arithmetic under stated assumptions only and makes no claim about any syllabus, board, class or real-world productivity.

Related reading on this site: how to solve maths word problems, the Working with Fractions chapter page (rates are fractions of a job), and maths tuition at EZ Tuitions.